Answer
a) $v_B =\dfrac{d}{m}\sqrt{ k(m+M) }$
b) $200.2\;\rm m/s$
c) $99.7\%$
Work Step by Step
a) We chose the system to be the bullet plus the block. So the momentum is conserved just before and just after the collision.
$$p_{ix}=p_{fx}$$
We know that the bullet and the block will move as one unit after the collision.
$$mv_B=(m+M)v_{fx}$$
Thus,
$$v_{fx}=\dfrac{mv_B}{m+M}\tag1$$
Now we need to find the compressed distance $d$ by the spring when the block with the bullet stops.
We consider the system here to be the bullet plus the block and the spring as an isolated system since the table is frictionless. f
$$E_i=E_f$$
$$U_{is}+K_i=U_{fs}+K_f$$
$$0+K_i=U_{fs}+0$$
where the initial here refers to the moment just before the collision, and the final refers to when the system stops. So, $v_i$ here is $v_{fx}$
$$ \color{red}{\bf\not} \frac{1}{2}(m+M)v_{fx}^2=\color{red}{\bf\not} \frac{1}{2}kd^2 $$
$$ (m+M)v_{fx}^2= kd^2 $$
Thus,
$$v_{fx}=\sqrt{\dfrac{kd^2}{m+M}}=\sqrt{\dfrac{k}{m+M}}d$$
Plugging from (1);
$$ \dfrac{mv_B}{m+M}=\sqrt{\dfrac{k}{m+M}}d$$
$$v_B =\sqrt{\dfrac{k}{m+M}}d\times \dfrac{m+M}{m }$$
$$v_B =\dfrac{d}{m}\sqrt{\dfrac{k(m+M)^2}{m+M}} $$
$$\boxed{v_B =\dfrac{d}{m}\sqrt{ k(m+M) }} $$
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b) Here we just need to plug the given to find the speed of the bullet.
$$v_B =\dfrac{0.1}{0.005}\sqrt{ 50(0.005+2) }=\color{red}{\bf 200.2}\;\rm m/s$$
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c) The loss of energy is given by subtracting the final kinetic energy of the bullet plus the block as one unit from the initial kinetic energy of the bullet.
This loss of energy is due to the heat energy due to the friction between the bullet and the block plus the sound energy due to the collision.
$$E_{loss}=\dfrac{K_i-K_f}{K_i}=\dfrac{\frac{1}{2}mv_B^2-\frac{1}{2} (m+M)v_{fx}^2}{\frac{1}{2}mv_B^2}$$
Plugging from (1);
$$E_{loss} =\dfrac{\frac{1}{2}mv_B^2-\frac{1}{2} (m+M)\left[\dfrac{mv_B}{m+M}\right]^2}{\frac{1}{2}mv_B^2}$$
$$E_{loss} =\dfrac{\color{red}{\bf\not} \frac{1}{2}m\color{red}{\bf\not}
v_B^2- \dfrac{\color{red}{\bf\not} \frac{1}{2} m^2\color{red}{\bf\not}
v_B^2}{m+M} }{\color{red}{\bf\not} \frac{1}{2}m\color{red}{\bf\not}
v_B^2}$$
$$E_{loss} =\dfrac{ m - \dfrac{ m^2 }{m+M} }{ m }$$
Plugging the known;
$$E_{loss} =\dfrac{ 0.005- \dfrac{ 0.005^2 }{0.005+2} }{ 0.005}=\bf 0.997=\color{red}{\bf99.7}\bf\%$$
As we see, the bullet loses about 99.7$\%$ of its energy due to the heat energy produced by the friction force between the bullet and the block, and the sound energy due to the collision.
